Download Interferometry Lab: Michelson and Fabry-Perot, Schedules, Exam II, and Job Opportunity and more Study notes Optics in PDF only on Docsity!
Plans for tonight, and next 2 weeks
Labs tonight and Tuesday in Room B260, Michelson interferometer
group M-Z has priority tonight, A-L on Tuesday
Thursday 5/8: Fabry-Perot interferometry (both groups)
Thursday 5/13: Exam II (material covered since last exam, in
Chapters 19, plus
chapters 4,5,7 and 8)
Homework 2 solutions will posted at the end of next week, papers returned followingweek
I will have office hours Thursday 5/8 before and after class
I will be away all of the week of 5/12~5/16 – exam will be proctored by B. Feldman
Same format and procedure as Exam I: allow 1 week for grading
Sorry, no class Thursday 5/15 (2 weeks from tonight)
Job opportunity FYI:
From: Jeffrey Pearce Subject: FT TEMP ENGINEERING INSTRUCTOR JOB AVAILABLEEverett Community College, a growing 2-year education leader inWashington State, is searching for a highly-motivated, excellentteacher to serve as a full-time temporary Engineering Instructor.Salary is $43,946-$46,746.Please see this link for complete information:http://www.everettcc.edu/ftf
Visibility of fringes
Coherence length
length of wavetrains emitted by source
for thermal lamps, L
C
< mm
for lasers, L
C
~ meters or km
For path difference
= 2d > L
C
no fringes are seen
can’t have interference between separate wavetrains
Note: Fourier analysis says L
C
~ 1/(spectral width)
Truly monochromatic
→
L
C
~
∞
Broad spectrum
→
short L
C
So monochromaticity of source
∝
L
C
Fringes disappear for d>>L
C
because different
λ
overlap
Recall:
Suppose we add equal (E
i =E
0
) waves of many different
λ
:
L
C
cos
(^20)
δ
E
I
:
of
function
a
as
axis
on
Intensity
difference
Phase
difference
Path
1
2
1
2
2
2
1
2 1
1
2
As with random phases,incoherent cross terms average to 0:For a continuous
distribution:
2 cos(
j
i
i
t
i
t
i
t
i
t
j
j
i
j
E
E e
E e
I
EE
E
e
e
I
I
I
d
I
E
N
ω
ω
ω
ω
ω
ω
−
∑
∑
L
L
Fourier spectroscopy:
Can infer spectrum
from I(
(variation in V with
Examples:
Δ
Intensity
Δ
Intensity
Δ
Intensity
σ
Intensity
σ
Intensity
σ
Intensity
monochromatic 2 closely-spaced lines broad spectrum
constant V (V=1)
beats
V
→
0 (fringes disappear) for large
Δ
I(
) vs
I(
σ
) vs
σ
Multiple-beam interferometry: Fabry-Perot
Fabry-Perot interferometer (C. Fabry & A. Perot, 1897)
Multiple amplitude division
2 partially-reflecting surfaces with (small) spacing t
Fixed-spacing modules used as filters are called “etalons”
Mirrors are characterized by (intensity) reflectance R
Reflected intensity I
R
=RI
0
R=r
2
where r=
amplitude reflectance factor: E
R
=rE
0
=
√
R E
0
)
Typically assume T+R=1 (but there can be absorption also)
Geometric path difference:
E
0
E
1
E
2
E
3
E
4
E
0
E
(^1) E
2
l
l
l
d
t
θ
θ
θ
Multiple beams
Geometric path difference
θ = θ − θ = θ
θ
θ
θ = θ = θ = − = Δ
cos
sin
cos
cos
sin
cos
tan
sin
cos
2
2
21
21
t
t
t
t
t d d p t l p l
So
and
but
t
F-P interferometry
Combine separate rays (use lens or view from
1 Then
2
(
)
2
(
2
)
3
(
3
)
0
2
2
3
3
1
(
0
(^20)
Note: So
NET
i
t
i
t
i
t
i
t
NET
i
t
i
i
i
N
i N
NET
N
i
t
NET
i
E
E
E
E
E
Te
RTe
R Te
R Te
E
E Te
R e
R e
R e
R
e
x
x
x
x
E Te
E
R e
ω
ω
δ
ω
δ
ω
δ
ω
δ
δ
δ
δ
ω
δ
−
−
−
−
−
−
−
−
−
−
L
L
L
L
E
0
E
1
E
2
E
3
E
4
T
1/
R
1/
T
1/
T
1/
R
1/
R
1/
R
1/
R
1/
(
) (
)
1
2
2
0
2 1
2
2
0
2
(^21)
2
2
0
2 2
1
Re[
]
2
1
1
1 1
2
cos
i
i
i
i
I
EE
E
T
R e
R e
E
T
I
R
R e
R e
E
T
R
R δ
δ
δ
δ
δ
−
−
= =
−
−
=
−
−
=
−
Fabry-Perot interferometry
Since I
0
= (1/2)E
0
2
, we have
For fixed R, I=I(
δ
), max occurs when sin
2
term=
Minimum occurs when sin
2
term=
So
(notice:
independent of T)
2
0
2
2
2
2
2
2
2
2
(
phase diff between adjacent beams)
Denominator
but
So denominator
cos
cos
cos
cos
cos
2 sin
sin
I T
I
R
R
R
R
R
R
R
R
x
R
R
x
R
R
R
δ
⎛
⎞
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎛
⎞
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
=
=
2
2 2
2
0
2
2
2
2
(Airy's formula)
sin
sin
R
R
I T
I
R
R
R
δ
δ
⎛
⎞
⎜
⎟
⎝
⎠
⎛
⎞
⎜
⎟
⎝
⎠
2
2
0
)
1 (
)
2 (
R T
I
I
m
MAX
MAX
− = → π = δ
2
2
0
2
2
0
R
T
I
R
R
T
I
I
m
MIN
MIN
2 2
MAX MIN
I
R
I
R
Resolving power of FP
Fringe width
FWHM (full width at half-max)
HWHM=FWHM/2 (half width)
m
0
=order # at center
Let dm=HWHM.
Then for T=1-R:
At maximum, m
0
=even integer, so
sin(
π
(m
0
+dm))=sin(
π
dm)
(periodicity of sine function)
Intensity
m
fringe order #
dm
m
t
m
π
δ
θ
λ
cos
2
0
2
2
2
2 2
0
2
define
But when
sin
sin
sin
MAX
MAX
MAX
R
F
R
I
I
I
I
F
R
R
I
I
I
m
m
dm
I
F
m
δ
δ
⎛
⎞
⎛
⎞
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
FWHM
(
)
(
)
sin
sin
2
2
dm
F
dm
F
I
I
MAX
is
max
half
for
condition
so
Finesse and resolution
If R is large, F>>1, so this condition requires sin
2
factor small:
Define
finesse
F
=(separation of orders)/(FWHM of fringe)
by definition, separation of orders=1 unit on m axis, so
F
=1000 is achievable
Note common sources of confusion:
– F=“
coefficient of finesse”
(Fabry)
F
=just plain “finesse” )
- Many books use the term “halfwidth” to mean FWHM
F
dm
FWHM
F
dm
dm
F
dm
dm
sin(
2
and
F
dm
F
Taylor Criterion
But for line 1
at the same point we have 2tcos
θ
= (m
1
λ
Example: lab setup next week has R=0.5, t=0.001 m
then
m
Intensity
dm
m
1
m
2
order m or
λ
cos
2
0
1 2 R^ R
t
d
R
R F F t m d
θ
t
m
θ
m
F
m
d
F
dm
dm
F
dm
I
I
dm
m
d
dm
m
d
m
T
λ
π
λ λ
− = λ π = = λ
λ
λ = = π = λ
λ
π
π
λ λ → λ + = λ + λ
←
←
that
Recall
axis)
(on
For
(CRP)
Power
Resolving
Chromatic
:
is
max)
half
at
(lines
criterion
Taylor
So
when
that
found
just
we
But
F
F
cross
14000
)
5 .
0
1
)(
7
10
3 .
6 (
5 .
0
)
001 .
0 (
2
≈
−
−
π ×
=
λ λ
m m
d
so for
λ
~600 nm, we can resolve
0.04 nm separation!
